Finite Fields and Their Applications 13 (2007) 1071–1085

http://www.elsevier.com/locate/ffa

Polyadic codes of prime power length

Anuradha Sharma 1, Gurmeet K. Bakshi, Madhu Raka ∗

Centre for Advanced Study in Mathematics, Panjab University, Chandigarh 160014, India

Received 31 January 2006; revised 19 November 2006

Available online 17 December 2006

Communicated by W. Cary Huffman

Dedicated to the memory of Professor V.C. Dumir

Abstract

Polyadic codes constitute a special class of cyclic codes and are generalizations of quadratic residuecodes, duadic codes, triadic codes, m-adic residue codes and split group codes, which have good error-correcting properties. In this paper, we give necessary and sufficient conditions for the existence of polyadiccodes of prime power length. Examples of some good codes arising from the family of polyadic codes ofprime power length are also given.© 2006 Elsevier Inc. All rights reserved.

Keywords: Cyclic codes; Cyclotomic cosets; Idempotents

1. Introduction

Let Fq be a finite field with q elements and N be a positive integer coprime to q . A cycliccode C of length N over Fq is an ideal in the principal ideal ring RN = Fq [x]/〈xN − 1〉. Theideal C is generated by a unique monic polynomial g(x), which is a divisor of (xN − 1), calledthe generator polynomial of C. For α, a fixed primitive N th root of unity in some extension fieldof Fq , the set {i (modulo N) | g(αi) = 0} is called the defining set of C.

For a positive integer m � 2, an m-adic splitting of Z/NZ is an (m + 1)-tuple (X∞,X0,X1,

. . . ,Xm−1) such that

* Corresponding author.E-mail address: mraka@pu.ac.in (M. Raka).

1 Research supported by N.B.H.M., India, is gratefully acknowledged.

1071-5797/$ – see front matter © 2006 Elsevier Inc. All rights reserved.doi:10.1016/j.ffa.2006.12.006

1072 A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085

1. each of X∞,X0,X1, . . . ,Xm−1 is a union of q-cyclotomic cosets modulo N ;2. X∞,X0,X1, . . . ,Xm−1 form a partition of Z/NZ, i.e.,

Z/NZ = X∞ ∪ X0 ∪ X1 ∪ · · · ∪ Xm−1

as a disjoint union; and3. there exists an integer a, gcd(a,N) = 1 such that μa(X∞) = X∞ and μa(Xi) = Xi+1 for

0 � i � m − 1, the subscripts are taken modulo m, where μa (called multiplier) denotes thepermutation i �→ ai (mod N).

It is clear that 0 ∈ X∞ always. Let X′∞ = X∞ − {0}. The four families of codes having(X′∞ ∪ Xi)

c , X∞ ∪ Xi , X′∞ ∪ Xi , and (X∞ ∪ Xi)c , for 0 � i � m − 1, as their defining sets, are

called polyadic codes, where Ac denotes the complement of the set A in Z/NZ.Thus the existence of polyadic codes of length N is equivalent to the existence of an m-adic

splitting of Z/NZ.In the special case, when m = 2 and X∞ = {0}, polyadic codes are duadic codes as defined

and studied extensively by Leon, Masley and Pless [9], Ding and Pless [4], Smid [15] and Ding,Lam and Xing [6], etc. When m = 3, polyadic codes are triadic codes as defined by Pless andRushanan [12]. When X∞ = {0}, N = p an odd prime and X0 is the set of all m-adic residuesmodulo p, polyadic codes are m-adic residue codes defined by Brualdi and Pless [3], and inves-tigated further, by V.R. Job [8]. Ding, Kohel and Ling [5] generalized duadic codes to split groupcodes, where the underlying group was taken to be any finite Abelian group. Brualdi and Pless [3]defined polyadic codes or m-adic codes and gave a necessary and sufficient condition for the ex-istence of an m-adic splitting of Z/pZ, where p is a prime. Ling and Xing [10] extended thedefinition of polyadic codes to include non-cyclic Abelian codes.

When N = pn, a prime power, an m-adic splitting of Z/pnZ was called ‘non-degenerate’

by Ling and Xing [10], if the set {λpn−1 | 0 � λ � p − 1} �⊆ X∞, i.e., Xj ∩ {λpn−1 | 0 � λ �p − 1} �= ∅ for all j , 0 � j � m − 1. They investigated the existence of ‘non-degenerate’ m-adicsplitting of Z/pn

Z.The constraint that an m-adic splitting of Z/pn

Z is ‘non-degenerate,’ considered by Ling andXing [10], is a very strong restriction.

This constraint prohibits p to be 2, as a ‘non-degenerate’ splitting of Z/2nZ does not exist

because the set {0,2n−1} is always contained in X∞.Splittings with this constraint do not always give rise to non-degenerate codes (a non-

degenerate code is a code which is not a repetition of a code of smaller block length). Forevery odd prime p, one can construct many examples of splittings with this constraint of ‘non-degeneracy,’ which give rise to degenerate codes. There are also many examples of splittingswithout this constraint of ‘non-degeneracy,’ which give rise to non-degenerate codes.

When p is an odd prime, the constraint of ‘non-degeneracy’ restricts m to have fewer choices.With this constraint, m has to be a factor of e only. However, without this constraint, we provethat m can be any factor of epd−1, where e = p−1

f, f is the multiplicative order of q modulo p

and qf = 1 + pdc, d � 1, with c not divisible by p.Therefore it is very important to find a necessary and sufficient condition for the existence

of an m-adic splitting of Z/pnZ, where p is any prime, even or odd, without the constraint of

‘non-degeneracy,’ which we investigate in this paper (see Theorems 1–4, Section 3). In Section 4,we give examples of some good codes arising from the family of polyadic codes of prime powerlength.

A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085 1073

In another paper [1], the authors explore the necessary and sufficient conditions for the exis-tence of polyadic codes of arbitrary length in terms of solvability of certain linear Diophantineequations.

2. q-Cyclotomic cosets

In this section, we determine all the q-cyclotomic cosets modulo pn, where p is a prime andn is a positive integer, which are needed in the proofs of Theorems 1–4.

A q-cyclotomic coset Cs of s in Z/NZ is the set {s, sq, sq2, . . . , sqNs−1}, where Ns is theleast positive integer such that sqNs ≡ s (mod N).

2.1. q-Cyclotomic cosets modulo an odd prime power

Let p be an odd prime and q be a prime power with gcd(p, q) = 1. Let Ok(q) denote themultiplicative order of q modulo k. Let Op(q) = f = p−1

efor some positive integer e and write

qf = 1 + pdc, where p does not divide c and d is a positive integer. For every positive integer �,

Op�(q) ={

fp�−d if � � d + 1,

f if � � d.(1)

Let g be a primitive root modulo p. On replacing g by g +p (if necessary), we can ensure that

gcd(gp−1−1

p,p) = 1, so that g is a primitive root modulo p� for every positive integer �. In view

of (1), the q-cyclotomic coset modulo pn containing pn−�gk for any �, 1 � � � n, and k � 0, isgiven by

Cpn−�gk ={ {pn−�gk,pn−�gkq, . . . ,pn−�gkqfp�−d−1} if d + 1 � � � n,

{pn−�gk,pn−�gkq, . . . ,pn−�gkqf −1} if 1 � � � d.(2)

Proposition 1. All the distinct q-cyclotomic cosets modulo pn are given by C0 = {0} and Cpn−�gk

for 1 � � � n and k runs over R� for each �, where

R� ={ {0,1,2, . . . , epd−1 − 1} if d + 1 � � � n,

{0,1,2, . . . , ep�−1 − 1} if 1 � � � d.

Proof. Clearly, for any k, k′, Cpn−�gk �= Cpn−�′gk′ if � �= �′. As g is a primitive root modulo p�,

one can easily check that Cpn−�gk �= Cpn−�gk′ for any �,1 � � � n, and k, k′ ∈ R�. Further, these

are all the q-cyclotomic cosets modulo pn because

|C0| +n∑

�=1

∑k∈R�

|Cpn−�gk | = pn. �

1074 A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085

2.2. q-Cyclotomic cosets modulo 2n, n a positive integer

Since q ≡ ±1 (mod 4), we write q = 1 + 2dc or −1 + 2dc, where d � 2 and c is odd. Notethat if q ≡ 3 or 5 (mod 8), d is always equal to 2. For every positive integer �, � � 2,

O2� (q) =⎧⎨⎩

2�−d if � � d + 1, q = ±1 + 2dc,

1 if 2 � � � d, q = 1 + 2dc,

2 if 2 � � � d, q = −1 + 2dc.

(3)

Therefore the q-cyclotomic coset modulo 2n containing 2n−�r , 2 � � � n, r odd, is given by

C2n−�r =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

{2n−�r,2n−�rq, . . . ,2n−�rq(2�−d−1)} if d + 1 � � � n

and q = ±1 + 2dc,

{2n−�r} if 2 � � � d

and q = 1 + 2dc,

{2n−�r,2n−�rq} = {2n−�r,−2n−�r} if 2 � � � d

and q = −1 + 2dc.

(4)

Clearly C0 = {0}, C2n−1 = {2n−1}.

Proposition 2.

(a) Let q = 1 + 2dc, d � 2, c odd. All the distinct q-cyclotomic cosets modulo 2n are givenby C0, C2n−1 and C2n−�r for 2 � � � n and r runs over S� for each �, where

S� ={

{±1,±3, . . . ,±3(2d−2−1)} if d + 1 � � � n,

{±1,±3, . . . ,±3(2�−2−1)} if 2 � � � d.

(b) Let q = −1 + 2dc, d � 2, c odd. All the distinct q-cyclotomic cosets modulo 2n are givenby C0, C2n−1 , C2n−2 and C2n−�r for 3 � � � n and r runs over T� for each �, where

T� =

⎧⎪⎨⎪⎩

{1,3,32, . . . ,3(2d−1−1)} if d + 1 � � � n, d � 3,

{1,3,32, . . . ,3(2�−2−1)} if 3 � � � d, d � 3,

{1,−1} if 3 � � � n, d = 2.

Proof. We first note that O2� (3) = 2�−2 for � � 3 and q2 ≡ 1 (mod 2d+1) for any q . Also forany r, r ′ odd, it is clear that C2n−�r �= C2n−�′ r ′ if � �= �′.

Case (a). q = 1 + 2dc, d � 2, c odd.Suppose, if possible, C2n−�r = C2n−�r ′ for some �,2 � � � n, and r, r ′ ∈ S�. This means

2n−�3t ≡ ±2n−�3t ′qu(mod 2n

)for some t , t ′, 0 � t, t ′ � 2min(�−2,d−2) −1 and for some u. Then 3t−t ′ ≡ ±qu (mod 2�). This im-plies 32(t−t ′) ≡ q2u ≡ 1 (mod 2min(�,d)+1). Since the order of 3 modulo 2min(�,d)+1 is 2min(�,d)−1,we get that 2min(�,d)−1 divides 2(t − t ′), which is possible only if t = t ′. Further if r = −r ′, weget 1 ≡ −qu (mod 2�), which is not possible for q = 1 + 2dc. Thus the cosets C2n−�r , 2 � � � n

A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085 1075

and r ∈ S�, are disjoint modulo 2n. Further these are all the q-cyclotomic cosets modulo 2n,because

|C0| + |C2n−1 | +n∑

�=2

∑r∈S�

|C2n−�r | = 1 + 1 +n∑

�=d+1

2d−12�−d +d∑

�=2

2�−1 = 2n.

Case (b). q = −1 + 2dc, d � 2, c odd.For d = 2, one finds easily that the cosets C2n−�r , r ∈ T� = {1,−1} for 3 � � � n, are disjoint

modulo 2n.Let now d � 3. Suppose, if possible, C2n−�r = C2n−�r ′ for some �, 3 � � � n and r, r ′ ∈ T�.

This gives

3t−t ′ ≡ qu(mod 2�

)(5)

for some t , t ′, 0 � t ′ � t � 2min(�−2,d−1) − 1 and for some u.For d + 1 � � � n, working as above, we get that it is possible only if either t − t ′ = 0 or

t − t ′ = 2d−2. If t − t ′ = 2d−2, write 3t−t ′ = 32d−2 = 1 + 2dλ, λ odd. Then (5) gives 3t−t ′ =1 + 2dλ ≡ qu (mod 2d+1), which is not possible as qu ≡ 1 (mod 2d+1) if u is even and qu ≡−1 (mod 2d) if u is odd. Thus we must have t = t ′.

For 3 � � � d , we get, from (5), that 32(t−t ′) ≡ q2u ≡ 1 (mod 2�+1). This implies 2�−1 divides2(t − t ′), which gives t = t ′, as t − t ′ < 2�−2. Therefore the cosets C2n−�r , 3 � � � n and r ∈ T�,are disjoint modulo 2n. Further these are all the q-cyclotomic cosets modulo 2n, because thenumber of elements in the listed cosets sums upto to 2n. �3. Existence of an m-adic splitting of ZZZ/pn

ZZZ

By a non-trivial m-adic splitting of Z/NZ, we mean an m-adic splitting (X∞,X0,X1, . . . ,

Xm−1) with Xj �= ∅ for each j , 0 � j � m − 1.For a fixed positive integer � and a prime p, let G� denotes the multiplicative group of reduced

residue system modulo p� of order φ(p�). Let Q� = 〈q〉 be a subgroup of G�, generated by q ,of order |Q�| = Op�(q).

By an m-adic splitting of G�, we mean an (m + 1)-tuple (Y∞, Y0, Y1, . . . , Ym−1) such that

1. each of the sets Y∞, Y0, Y1, . . . , Ym−1 is a union of cosets of Q� in G�;2. G� = Y∞ ∪ Y0 ∪ Y1 ∪ · · · ∪ Ym−1, as a disjoint union; and3. there exists an integer a, gcd(a,p) = 1 such that μa(Y∞) = Y∞ and μa(Yi) = Yi+1 for

0 � i � m − 1, where the subscripts are taken modulo m.

Lemma 1. Z/pnZ admits an m-adic splitting (X∞,X0,X1, . . . ,Xm−1) by a multiplier μa if

and only if G� for each �, 1 � � � n, admits an m-adic splitting (X(�)∞ ,X

(�)0 ,X

(�)1 , . . . ,X

(�)m−1) by

the same multiplier with |X′∞| = ∑n�=1 |X(�)∞ |. Further, if the splitting of Z/pn

Z is non-trivial,

then the m-adic splitting of G� is non-trivial for at least some � (i.e., |X(�)∞ | < φ(p�) for at leastone �).

1076 A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085

Proof. We have

Z/pnZ − {0} = pn−1G1 ∪ pn−2G2 ∪ · · · ∪ pGn−1 ∪ Gn.

Let (X∞,X0,X1, . . . ,Xm−1) be an m-adic splitting of Z/pnZ by a multiplier μa . As

gcd(a,p) = 1, we have μa(pn−�G�) = pn−�G� for every �, 1 � � � n. So each of the sets

pn−�G� must split separately. We note that the q-cyclotomic coset Cpn−�r with gcd(r,p) = 1, asdefined in Section 2, is the same as the set pn−�rQ� modulo pn for each �, 1 � � � n. Thereforeif we write

X′∞ =n⋃

�=1

pn−�X(�)∞ and Xj =n⋃

�=1

pn−�X(�)j for 0 � j � m − 1,

then |X′∞| = ∑n�=1 |X(�)∞ | and (X

(�)∞ ,X(�)0 ,X

(�)1 , . . . ,X

(�)m−1) is an m-adic splitting of G� for

every �, given by the same multiplier μa . The converse is clear. �For any a ∈ G�, let H

(�)a denote the subgroup of G�/Q�, generated by the coset aQ�, with

order O(H(�)a ) = α

(�)a and index [G�/Q� : H(�)

a ] = β(�)a . Then we have

φ(p�

) = O(H(�)

a

)|Q�|β(�)a . (6)

Proposition 3. For a prime p and a positive integer n, Z/pnZ admits a non-trivial m-adic

splitting (X∞,X0,X1, . . . ,Xm−1) with |X′∞| = t given by the multiplier μa if and only if

(i) t = t1 + t2 + · · · + tn, where 0 � t� � φ(p�) for each �, 1 � � � n, and t� < φ(p�) for atleast one �;

(ii) O(H(�)a )|Q�| divides t� for each �; and

(iii) m divides O(H(�)a ) for all those � for which t� < φ(p�).

Proof. Let (X∞,X0,X1, . . . ,Xm−1) be a non-trivial m-adic splitting of Z/pnZ given by the

multiplier μa . Then by Lemma 1, (X(�)∞ ,X

(�)0 ,X

(�)1 , . . . ,X

(�)m−1) is an m-adic splitting of G� for

every �. Condition (i) follows, on taking |X(�)∞ | = t�. Let

G�/Q� = b1H(�)a ∪ b2H

(�)a ∪ · · · ∪ b

β(�)a

H (�)a .

Since μa(biH(�)a ) = biH

(�)a , μa(X

(�)∞ ) = X(�)∞ and μa(X

(�)j ) = X

(�)j+1, each set biH

(�)a is either

contained in X(�)∞ or it splits equally into m parts, which lie in X

(�)0 ,X

(�)1 , . . . ,X

(�)m−1. Therefore

X(�)∞ is either empty or is a union of some of the sets biH

(�)a . In any case, O(H

(�)a )|Q�| must

divide t� = |X(�)∞ |, as each coset in H(�)a has |Q�| elements in it. This proves condition (ii).

Further, if t� < φ(p�), at least one biH(�)a is not contained in X

(�)∞ , so it has to split equally intom parts in such a way that each part must lie in exactly one of X

(�)0 ,X

(�)1 , . . . ,X

(�)m−1, therefore m

must divide |biH(�)a |, which is the same as O(H

(�)a ). This proves condition (iii).

A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085 1077

Conversely, suppose that (i)–(iii) hold. Let t� = O(H(�)a )|Q�|s�, where, by (6), 0 � s� � β

(�)a .

If s� < β(�)a for some �, i.e., if t� < φ(p�), we have, by (iii), that m | α(�)

a . Write

H(�)a = {

Q�,aQ�, a2Q�, . . . , a

α(�)a −1Q�

} = K(�)a ∪ aK(�)

a ∪ · · · ∪ am−1K(�)a ,

where K(�)a = {Q�,a

mQ�,a2mQ�, . . . , a

m(α(�)am

−1)Q�}. Take

X(�)∞ = b1H(�)a ∪ b2H

(�)a ∪ · · · ∪ bs�H

(�)a .

For each j,0 � j � m − 1, take

X(�)j =

{bs�+1a

jK(�)a ∪ bs�+2a

jK(�)a ∪ · · · ∪ b

β(�)a

ajK(�)a if s� < β

(�)a ,

∅ if s� = β(�)a .

Then (X(�)∞ ,X

(�)0 ,X

(�)1 , . . . ,X

(�)m−1) gives a partition of G� for every �, with μa(X

(�)∞ ) =X

(�)∞ ,μa(X(�)j ) = X

(�)j+1 and |X(�)∞ | = t�. Therefore, by Lemma 1, we get a non-trivial m-adic

splitting of Z/pnZ with |X′∞| = ∑n

�=1 t� = t , given by the multiplier μa . �We now distinguish the cases, when p is an odd prime or an even prime.

3.1. In this subsection, we take p to be always an odd prime.Following the notations of Section 2.1, G� here is a cyclic group generated by g, and G�/Q�

is also a cyclic group generated by the coset gQ�, where g is a primitive root modulo p�. FromProposition 1, we have

G� ={

Q� ∪ gQ� ∪ g2Q� ∪ · · · ∪ gepd−1−1Q� if � � d + 1,

Q� ∪ gQ� ∪ g2Q� ∪ · · · ∪ gep�−1−1Q� if 1 � � � d.(7)

Therefore

O(G�/Q�) = epmin(�−1,d−1). (8)

For any integer a with gcd(a,p) = 1, we note that H(�)a = 〈aQ�〉 = 〈gsQ�〉 for some s, 0 �

s � min(ep�−1 − 1, epd−1 − 1). Write s = s′pr , where 0 � r � min(� − 1, d − 1). Therefore

O(H(�)

a

) = e

gcd(s′, e)pmin(�−1−r,d−1−r). (9)

Also

O(H(�)

a

)|Q�| = ef

gcd(s′, e)p�−1−r . (10)

Firstly, we note that if Z/pnZ admits a non-trivial m-adic splitting, then by Proposition 3,

m must divide O(H(�)a ) for at least one �. Hence, by (9), we obtain that m divides epd−1. Let

m = m′pλ, where m′ | e and λ � d − 1. We consider the two cases, m = m′pd−1 and m = m′pλ,0 � λ � d − 2, separately.

1078 A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085

Theorem 1. Let p be an odd prime and m = m′pd−1, where m′ | e. Then Z/pnZ admits

a non-trivial m-adic splitting (X∞,X0,X1, . . . ,Xm−1) with |X′∞| = t , |X0| = |X1| = · · · =|Xm−1| = k if and only if

(i) pn − 1 = t + mk, k �= 0, where both t and k are divisible by f ,(ii) in the p-adic expansion of t = a0 + a1p + a2p

2 + · · · + an−1pn−1, 0 � ai � p − 1, each ai

is divisible by m′f .

In fact, the conditions (i) and (ii) above are also necessary and sufficient for the existence ofa non-trivial m-adic splitting of Z/pn

Z for any m = m′pλ, m′ | e, 0 � λ � d − 1, if we restrictthat the multiplier is of the form μgs with gcd(s,p) = 1.

Proof. Suppose that (X∞,X0,X1, . . . ,Xm−1) is a non-trivial m-adic splitting of Z/pnZ given

by a multiplier μa . As each of X′∞,X0,X1, . . . ,Xm−1 is a union of non-zero q-cyclotomic cosetsmodulo pn and each such coset has cardinality divisible by f , by (2); so f must divide t and k.Moreover pn − 1 = t + mk as (X∞,X0,X1, . . . ,Xm−1) is a partition of Z/pn

Z.By Proposition 3, t = ∑n

�=1 t�, where 0 � t� � φ(p�). If t� < φ(p�), then by Proposition 3(ii),

m = m′pd−1 must divide O(H(�)a ). From (9), m will divide O(H

(�)a ) only if d − 1 � min(� −

1 − r, d − 1 − r) and m′ | egcd(s′,e) . This will happen only when r = 0. Also by Proposition 3(iii),

O(H(�)a )|Q�| divides t�. Therefore using (10), we get that ef

gcd(s′,e)p�−1 divides t�. Consequently,

m′fp�−1 divides t�. Write t� = a�−1p�−1, where 0 � a�−1 � p − 1 and m′f | a�−1. This proves

condition (ii).Conversely, suppose that (i) and (ii) hold. Since pn − 1 = t + m′pd−1k, k �= 0, we have

that n � d and pd−1 divides 1 + ∑d−2i=0 aip

i . But 1 � 1 + ∑d−2i=0 aip

i � pd−1, so we get t =pd−1 − 1 + ad−1p

d−1 +· · ·+ an−1pn−1. Also m′f divides ai for each i, so we have ai = m′f bi

for some bi , 0 � bi � em′ . Thus

t = φ(p) + φ(p2) + · · · + φ

(pd−1) + m′f bd−1p

d−1 + · · · + m′f bn−1pn−1.

Take

t� ={

φ(p�) if 1 � � � d − 1,

m′f b�−1p�−1 if d � � � n,

so that t = ∑n�=1 t�, where t� < φ(p�) for at least one �, as t < pn − 1. Let a = g

em′ . Then we

see, from (10), that O(H(�)a )|Q�| = m′fp�−1, which divides t� for each �. From (9), we have

O(H(�)a ) = m′pmin(�−1,d−1), which is divisible by m for every � � d . Hence, by Proposition 3,

there exists a non-trivial m-adic splitting (X∞,X0,X1, . . . ,Xm−1) of Z/pnZ with |X′∞| = t ,

given by μg

em′ . �

For λ � d − 2, if we do not want the restriction on the multiplier to be of the kind μgs

with gcd(s,p) = 1, then we can omit the condition (ii) of Theorem 1. In that case, we have thefollowing result:

A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085 1079

Theorem 2. Let p be an odd prime and m = m′pλ, where m′ | e and 0 � λ � d −2. Then Z/pnZ

admits a non-trivial m-adic splitting (X∞,X0,X1, . . . ,Xm−1) with |X′∞| = t and |X0| = |X1| =· · · = |Xm−1| = k if and only if pn − 1 = t + mk, k �= 0, with t and k both divisible by f .

Proof. Working as in Theorem 1, the given condition is clearly necessary for a non-trivial m-adicsplitting of Z/pn

Z to exist.Conversely, let pn − 1 = t + m′pλk, where k �= 0, f | t , f | k, m′ | e, 0 � λ � d − 2. First of

all, we must have n � λ + 1 and pλ | t + 1. Also pn−λ − 1 = t+1pλ − 1 + m′k. As m′f | m′k and

m′f | pn−λ −1, we must have m′f | t+1pλ −1. Write t+1

pλ −1 = m′f A, i.e., t = pλ −1+m′f Apλ,

where 0 � A <pn−λ−1

m′f .

Case I. 0 � A � p−1m′f = e

m′ . (Note that A < em′ if n = λ + 1.)

Here t = φ(p) + φ(p2) + · · · + φ(pλ) + m′f Apλ, where m′f Apλ � φ(pλ+1). Take

t� =⎧⎨⎩

φ(p�) if 1 � � � λ,

m′f Apλ if � = λ + 1,

0 if λ + 2 � � � n,

so that t = ∑n�=1 t�. Take a = g

em′ . Then, as in Theorem 1, O(H

(�)a )|Q�| = m′fp�−1 divides t�

for every �, and m = m′pλ divides O(H(�)a ) = m′pmin(�−1,d−1) for every � � λ + 1. Therefore,

by Proposition 3, there exists a non-trivial m-adic splitting of Z/pnZ with |X′∞| = t and |Xj | = k

by the multiplier μa .

Case II. em′ < A <

pn−λ−1m′f .

Here, we can write t = pλ − 1 + m′f Apλ = pλ − 1 + m′fpλ em′ + m′fpλ(A − e

m′ ) = pλ+1 −1 + m′f Bpλ, where

0 < B = A − e

m′ <pn−λ − 1 − ef

m′f= pn−λ − p

m′f. (11)

Let B = a0 + a1p + · · · + an−λ−1pn−λ−1, where 0 � ai � p − 1. All of a1, a2, . . . , an−λ−1

cannot be � em′ , because if so, then B � e

m′ (p + p2 + · · · + pn−λ−1) = pn−λ−pm′f , a contradiction

to (11). Let �0, 1 � �0 � n − λ − 1, be the largest index such that a�0 < em′ and ai � e

m′ for�0 + 1 � i � n − λ − 1. If any of a�0+1, a�0+2, . . . , an−λ−1 is > e

m′ , say aj � em′ + 1, then

B � ajpj + aj+1p

j+1 + · · · + an−λ−1pn−λ−1

�(

e

m′ + 1

)pj + e

m′ pj+1 + · · · + e

m′ pn−λ−1

= pn−λ + pj (m′f − 1)

m′f>

pn−λ − p

m′f,

a contradiction to (11). Hence, we must have a�0 < em′ and a�0+1 = a�0+2 = · · · = an−λ−1 = e

m′ .Thus we have

1080 A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085

t = φ(p) + φ(p2) + · · · + φ

(pλ+1) + a0m

′fpλ + a1m′fpλ+1 + · · · + a�0−2m

′fpλ+�0−2

+ a�0−1m′fpλ+�0−1 + a�0m

′fpλ+�0 + φ(pλ+�0+2) + φ

(pλ+�0+3) + · · · + φ

(pn

).

Take

t� =⎧⎨⎩

φ(p�) if 1 � � � λ + 1, λ + �0 + 2 � � � n,

a�−λ−2m′fp�−2 if λ + 2 � � � λ + �0,

a�0−1m′fpλ+�0−1 + a�0m

′fpλ+�0 if � = λ + �0 + 1.

Note that t�0+λ+1 < pm′fpλ+�0−1 + ( em′ − 1)m′fpλ+�0 = φ(p�0+λ+1).

Take a = gep

m′ . Then we get, from (10), that O(H(�)a )|Q�| = m′fp�−2, which divides t� for

every �. From (9), we have O(H(�)a ) = m′pmin(�−2,d−2), which is divisible by m = m′pλ for

λ+ 2 � � � λ+ �0 + 1 because λ � d − 2 here. Therefore, by Proposition 3, we get a non-trivialm-adic splitting of Z/pn

Z with |X′∞| = t and |Xj | = k given by the multiplier μa . �Remark 1. The result of Brualdi and Pless [3, Theorem 3.6] follows as a special case of Theo-rem 2 with n = 1.

3.2. In this subsection, we always take p = 2.If n = 1, clearly Z/2Z admits a trivial splitting only. If n = 2 and q = −1 + 2dc, d � 2, c

odd, then also, Z/4Z admits a trivial splitting only with X∞ = C0 ∪C1 ∪C2, where C1 = {1,3},C2 = {2}, as μa(Cs) = Cs for any a odd. If n = 2 and q = 1 + 2dc, c odd, the only non-trivialsplitting which Z/4Z can admit, is X∞ = C0 ∪ C2, X0 = C1 = {1}, X1 = C−1 = {3} with multi-plier μ−1 and m = 2. So let us take n � 3 from now onwards.

Theorem 3. Let q = 1 + 2dc, d � 2, c odd. Then Z/2nZ, n � 3, admits a non-trivial m-adic

splitting (X∞,X0,X1, . . . ,Xm−1) with |X′∞| = t , |X0| = |X1| = · · · = |Xm−1| = k if and only if

(i) 2n − 1 = t + mk, k �= 0;(ii) m = 2λ, where 1 � λ � d − 2 if d � 3, and λ = 1 if d = 2;

(iii) t �= 2λ − 1 if λ � 2.

Proof. Let (X∞,X0,X1, . . . ,Xm−1) be a non-trivial m-adic splitting of Z/2nZ. Clearly

2n − 1 = t + mk, k �= 0.By Lemma 1, each of G�, 1 � � � n, has an m-adic splitting (X

(�)∞ ,X(�)0 ,X

(�)1 , . . . ,X

(�)m−1)

with |X(�)∞ | = t�, and at least one G�, say G�0 , has a non-trivial m-adic splitting. We must have�0 � 2, as G1 can have trivial splitting only. For � � 2, by Proposition 2(a), we have

G� =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

Q� ∪ (3Q�) ∪ (32Q�) ∪ · · · ∪ (32�−2−1Q�)

∪ (−Q�) ∪ (−3Q�) ∪ · · · ∪ (−32�−2−1Q�) if 2 � � � d,

Q� ∪ (3Q�) ∪ (32Q�) ∪ · · · ∪ (32d−2−1Q�)

∪ (−Q�) ∪ (−3Q�) ∪ · · · ∪ (−32d−2−1Q�) if d + 1 � � � n.

(12)

Therefore

O(G�/Q�) = 2min(�−1,d−1). (13)

A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085 1081

First suppose that min(�0, d) � 3. As G�0 has a non-trivial splitting, we must have, by Propo-

sition 3, that m divides O(H(�0)a ). Here H

(�0)a is a cyclic subgroup of the non-cyclic group

G�0/Q�0 , therefore, O(H(�0)a ) must divide 1

2O(G�0/Q�0) = 2min(�0−2,d−2). Thus we have

m = 2λ, where 1 � λ � min(�0 − 2, d − 2). (14)

If �0 = 2 or d = 2, we have, from (12), that G�0 = Q�0 ∪ (−Q�0); so G�0 can have only 2-adicsplitting by the multiplier μ−1. Hence m = 2 in this case. This proves condition (ii).

Further, if λ � 2, min(�0, d) is at least 3. Then we have, from (14), that �0 � λ + 2, whichimplies that for � � λ + 1, G� can have trivial splitting only. This gives t� = φ(2�) for � � λ + 1.Thus t = ∑n

�=1 t� �∑λ+1

�=1 φ(2�) = 2λ+1 − 1 which gives condition (iii).Conversely, let (i)–(iii) hold.Case I. λ = 1, i.e., m = 2.Here 2n − 1 = t + 2k, k �= 0 and 2 divides t + 1. Write t+1

2 − 1 into the binary expansiont+1

2 − 1 = a0 +a12 +a222 +· · ·+an−22n−2 so that t = 1 +a02 +a122 +· · ·+an−22n−1, whereai = 0 or 1. Take

t� ={

1 if � = 1,

a�−22�−1 if 2 � � � n.(15)

Then t = ∑n�=1 t� and t� is zero for at least one �. From (12) and (3), we see that

O(H

(�)−1

)|Q�| ={

2 if 2 � � � d,

2�−d+1 if d + 1 � � � n.(16)

From (15) and (16), we find that O(H(�)−1 )|Q�| divides t� for every �, 1 � � � n (noting that

O(H(�)−1 )|Q�| = 1 if � = 1). Also m = 2 divides O(H

(�)−1 ) = 2 for every � � 2. Therefore, by

Proposition 3, Z/2nZ admits a non-trivial 2-adic splitting with |X′∞| = t by the multiplier μ−1.

Case II. λ � 2.By given hypotheses (i)–(iii), we have 2n − 1 = t + 2λk, k �= 0 and t �= 2λ − 1. Since 2λ

divides t + 1, we must have

0 � t + 1

2λ− 2 < 2n−λ − 2. (17)

Write t+12λ − 2 into its binary representation t+1

2λ − 2 = a0 +a12 +· · ·+an−λ−12n−λ−1, where

ai = 0 or 1. All of a1, a2, . . . , an−λ−1 cannot be 1, because otherwise t+12λ − 2 � 2 + 22 + · · · +

2n−λ−1 = 2n−λ − 2, a contradiction to (17). Let j , 1 � j � n − λ − 1, be the largest index suchthat aj = 0 and aj+1 = aj+2 = · · · = an−λ−1 = 1. Therefore t = φ(2)+φ(22)+· · ·+φ(2λ+1)+2λa0 + a12λ+1 + · · · + aj−12λ+j−1 + φ(2λ+j+2) + · · · + φ(2n). Let

t� ={

φ(2�) if 1 � � � λ + 1, λ + j + 2 � � � n,

a�−2−λ2�−2 if λ + 2 � � � λ + j + 1.

Thus t = ∑n�=1 t� with 0 � t� � φ(2�) and t� = a�−λ−22�−2 < φ(2�) for λ + 2 � � � λ + j + 1.

Take μa = μ3. From (12) and (3), we find that O(H(�)

)|Q�| = 2�−2, which divides t� for every �,

3

1082 A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085

and m = 2λ for 2 � λ � d − 2 divides O(H(�)3 ) = 2min(�−2,d−2) for every � � λ + 2. Therefore,

by Proposition 3, we get a non-trivial m-adic splitting of Z/2nZ with |X′∞| = t . �

Theorem 4. Let q = −1 + 2dc, d � 2, c odd. Then Z/2nZ, n � 3, admits a non-trivial m-adic

splitting (X∞,X0,X1, . . . ,Xm−1) with |X′∞| = t , |X0| = |X1| = · · · = |Xm−1| = k if and only if

(i) 2n − 1 = t + mk, k �= 0 and k even;(ii) m = 2λ, where 1 � λ � d − 1.

Proof. Let (X∞,X0,X1, . . . ,Xm−1) be a non-trivial m-adic splitting of Z/2nZ. Clearly

2n − 1 = t + mk, k �= 0. Further, since each of X0,X1, . . . ,Xm−1 is a union of non-zero q-cyclotomic cosets modulo 2n and, by (3), each such coset (except C2n−1 which must lie in X′∞)has cardinality divisible by 2, we get that 2 divides k. By Lemma 1, each of G�, 1 � � � n, hasan m-adic splitting and there exists some �0 such that G�0 has non-trivial m-adic splitting. SinceG1 and G2 cannot have a non-trivial splitting, we must have �0 � 3.

For � � 3, by Proposition 2(b), we have

G� =

⎧⎪⎨⎪⎩

Q� ∪ (3Q�) ∪ (32Q�) ∪ · · · ∪ (32�−2−1Q�) if 3 � � � d and d � 3,

Q� ∪ (3Q�) ∪ (32Q�) ∪ · · · ∪ (32d−1−1Q�) if d + 1 � � � n and d � 3,

Q� ∪ (−Q�) if d = 2 and � � 3.

(18)

Here

O(G�/Q�) ={

2min(�−2,d−1) if d � 3, 3 � � � n,

2 if d = 2, 3 � � � n.(19)

As G�0 has a non-trivial splitting, we must have, by Proposition 3, that m divides O(H(�0)a ).

But O(H(�0)a ) divides O(G�0/Q�0). Therefore, from (19), we get that m = 2λ, where 1 � λ �

d − 1.Conversely, let (i) and (ii) hold. Since 2n − 1 = t + 2λk, k �= 0 and 2 | k, we have 2λ+1 | t + 1.

Let t+12λ+1 − 1 = a0 + a12 + a222 +· · ·+ an−λ−22n−λ−2, i.e., t = φ(2)+φ(22)+· · ·+φ(2λ+1)+

2λ+1a0 + · · · + an−λ−22n−1, where ai = 0,1. Take

t� ={

φ(2�) if 1 � � � λ + 1,

2�−1a�−λ−2 if λ + 2 � � � n.

Then t = ∑n�=1 t�. If d � 3, take μa = μ3 and if d = 2, take μa = μ−1. One can easily check that

the conditions (ii) and (iii) of Proposition 3 are satisfied and hence, we get a non-trivial m-adicsplitting of Z/2n

Z with |X′∞| = t . �Remark 2. Given an m-adic splitting of Z/pn

Z, the idempotent generators of polyadic codesof length pn are also effectively computable. This is so, because the idempotent generators ofpolyadic codes of length pn can be explicitly written in terms of the idempotent generators ofminimal cyclic codes of length pn, making use of Theorem 4.3.7 of Huffman and Pless [7]. Theidempotent generators of binary minimal cyclic codes are easy to compute. For reference, see

A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085 1083

Van Lint [11, p. 52]. The idempotent generators of non-binary minimal cyclic codes of length pn,

p an odd prime, have been effectively computed by authors [13] when p does not divide qf −1p

if n � 2, where Op(q) = f . In [14], the authors have computed all the idempotent generators ofminimal cyclic codes of length 2n.

Table 1

Length pn q f m g Defining set Dimension Minimum distance

73 2 3 2 3 C1 ∪ Cg ∪ C7 ∪ C72 25 28

232 3 11 2 5 C1 ∪ Cg ∪ C23 12 184

372 3 18 2 2 C1 ∪ Cg ∪ C37 19 370

132 5 4 3 2 C1 ∪ Cg ∪ Cg2 ∪ C13 9 52

132 5 4 3 2 Cg ∪ Cg2 ∪ C13g ∪ C13g2 57 8

192 5 9 2 2 C1 ∪ Cg ∪ C19 10 133

292 5 14 2 2 C1 ∪ Cg ∪ C29 15 319

312 5 3 5 3 Cg2 ∪ C

g3 ∪ · · · ∪ Cg9

∪ C31g2 ∪ · · · ∪ C31g9

193 20

53 7 4 5 2 C1 ∪ Cg ∪ Cg2 ∪ C

g3

∪ Cg4 ∪ C5 ∪ C52

17 30

53 7 4 5 2 C0 ∪ Cg ∪ · · · ∪ Cg4

∪ C5g ∪ · · · ∪ C5g4

28 19

54 7 4 5 2 C1 ∪ Cg ∪ · · · ∪ Cg4

∪ C5 ∪ C5g ∪ · · · ∪ C5g4

∪ C52 ∪ C53

17 150

54 7 4 5 2 Cg ∪ Cg2 ∪ · · · ∪ C

g4

∪ C5g ∪ · · · ∪ C5g4

∪ C52g∪ · · · ∪ C52g4

129 19

54 7 4 5 2 C1 ∪ Cg ∪ Cg2 ∪ C

g3

∪ Cg4 ∪ C5 ∪ C52 ∪ C53

97 30

192 7 3 2 2 C1 ∪ Cg ∪ · · · ∪ Cg5

∪ C19 ∪ C19g2 ∪ C19g4

10 152

192 7 3 2 2 C1 ∪ Cg ∪ · · · ∪ Cg5

∪ C19 ∪ C19g ∪ · · · ∪ C19g3

7 190

192 7 3 2 2 C1 ∪ Cg ∪ Cg2 ∪ C

g3

∪ Cg4 ∪ C

g5 ∪ C19

16 57

292 7 7 4 2 Cg ∪ Cg2 ∪ C

g3 ∪ C29g

∪ C29g2 ∪ C29g3

211 19

292 7 7 2 2 C1 ∪ Cg ∪ Cg2

∪ Cg3 ∪ C29 ∪ C29g

15 232

292 7 7 4 2 C1 ∪ Cg ∪ Cg2 ∪ C

g3 ∪ C29 22 174

1084 A. Sharma et al. / Finite Fields and Their Applications 13 (2007) 1071–1085

4. Examples

Many different interesting examples arise from the family of polyadic codes of prime powerlength. In Table 1, we list some good polyadic codes of odd prime power length for q = 2,3,5and 7. The minimum distances of these codes have been computed with the help of com-puter algebra package ‘Magma.’ For all these polyadic codes, t , the cardinality of X′∞, hasthe p-adic expansion of the form a0 + a1p + a2p

2 + · · · , where each of the coefficients ai isdivisible by m′f . Below we discuss five polyadic codes with parameters [73,25,28] over F2,[529,12,184] over F3, [169,9,52] over F5, [125,17,30] over F7 and [16,3,12] over F7.

(i) For p = 7, n = 3, q = 2, we have f = 3, d = 1, e = 2, g = 3, m = m′ = 2. All the2-cyclotomic cosets modulo 73 are C0, C1, Cg , C7, C7g , C72 , C72g . For the splitting X′∞ =C1 ∪ Cg , X0 = C7 ∪ C72 , X1 = C7g ∪ C72g given by the multiplier μg , the polyadic code withdefining set X′∞ ∪ X0 has the parameters [73,25,28]. Here t = |X′∞| = 294 = a0 + a17 + a272

with a0 = a1 = 0, a2 = 6, each of these coefficients is divisible by m′f .(ii) For p = 23, n = 2, q = 3, we have f = 11, d = 1, e = 2, g = 5, m = m′ = 2. All the

3-cyclotomic cosets modulo 529 are C0, C1, Cg , Cp , Cpg . For the splitting X′∞ = C1 ∪ Cg ,X0 = Cp , X1 = Cpg given by the multiplier μg , the polyadic codes with defining sets X′∞ ∪ X0and X∞ ∪X0 have the parameters [232,12,184] and [232,11,207] respectively. Here t = 506 =a0 + a1p with a0 = 0, a1 = 22, both are divisible by m′f .

(iii) For p = 13, n = 2, q = 5, we have f = 4, d = 1, e = 3, g = 2, m = m′ = 3. All the5-cyclotomic cosets modulo 132 are C0, C1, Cg , Cg2 , C13, C13g , C13g2 . For the splitting withX′∞ = C1 ∪Cg ∪Cg2 , X0 = C13, X1 = C13g , X2 = C13g2 given by the multiplier μg , the polyadiccode with defining set X′∞∪X0 has the parameters [132,9,52]. Here t = |X′∞| = 156 = a0 +a1p

with a0 = 0, a1 = 12, both are divisible by m′f .(iv) For p = 5, n = 3, q = 7, we have f = 4, d = 2, e = 1, g = 2, m = 5, m′ = 1. All the

7-cyclotomic cosets are C0, C1, Cg , Cg2 , Cg3 , Cg4 , C5, C5g , C5g2 , C5g3 , C5g4 , C52 . For thesplitting X′∞ = C1 ∪ Cg ∪ Cg2 ∪ Cg3 ∪ Cg4 ∪ C52 , X0 = C5, X1 = C5g , X2 = C5g2 , X3 = C5g3 ,X4 = C5g4 given by the multiplier μg , the polyadic code with defining set X′∞ ∪ X0 has theparameters [53,17,30]. Here t = |X′∞| = 104 = a0 + a15 + a252 with a0 = 4, a1 = 0, a2 = 4,each of the coefficients is divisible by m′f .

(v) For p = 2, n = 4, q = 7, all the 7-cyclotomic cosets modulo 16 are C0, C1, C3, C9, C11,C2, C6, C4, C8. For the splitting X′∞ = C1 ∪C3 ∪C9 ∪C11 ∪C4 ∪C8, X0 = C2, X1 = C6, givenby the multiplier μ3, the polyadic code with defining set X′∞ ∪X0 has the parameters [24,3,12].The minimum distance of this code is as good as the best known in Brouwer’s table [2].

References

[1] G.K. Bakshi, M. Raka, A. Sharma, Existence of polyadic codes in terms of Diophantine equations, in: Proceedingsof International Conference of Diophantine Equations held at TIFR, Bombay, December 17–20, 2005, in honour ofProfessor T.N. Shorey’s 60th birthday, in press.

[2] A. Brouwer, Bounds on the minimum distance of linear code, available (on-line): http://www.win.tue.nl/~aeb/voorlincod.html.

[3] R.A. Brualdi, V.S. Pless, Polyadic codes, Discrete Appl. Math. 25 (1989) 3–17.[4] C. Ding, V. Pless, Cyclotomy and duadic codes of prime lengths, IEEE Trans. Inform. Theory 45 (2) (1999) 453–

466.[5] C. Ding, D.R. Kohel, S. Ling, Split group codes, IEEE Trans. Inform. Theory 46 (2) (2000) 485–495.[6] C. Ding, K.Y. Lam, C. Xing, Enumeration and construction of all duadic codes of length pm, Fund. Inform. 38 (1)

(1999) 149–161.

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[7] W.C. Huffman, V. Pless, Fundamentals of Error-Correcting Codes, Cambridge Univ. Press, 2003.[8] V.R. Job, M-adic residue codes, part I, IEEE Trans. Inform. Theory 38 (2) (1992) 496–501.[9] J.S. Leon, J.M. Masley, V. Pless, Duadic codes, IEEE Trans. Inform. Theory IT-30 (1984) 709–714.

[10] S. Ling, C. Xing, Polyadic codes revisited, IEEE Trans. Inform. Theory 50 (1) (2004) 200–207.[11] J.H. Van Lint, Coding Theory VII, vol. 201, Springer, Berlin, 1971.[12] V. Pless, J.J. Rushanan, Triadic codes, Linear Algebra Appl. 98 (1988) 415–433.[13] A. Sharma, G.K. Bakshi, V.C. Dumir, M. Raka, Cyclotomic numbers and primitive idempotents in the ring

GF(q)[x]/(xpn − 1), Finite Fields Appl. 10 (4) (2004) 653–673.[14] A. Sharma, G.K. Bakshi, V.C. Dumir, M. Raka, Irreducible cyclic codes of length 2n, Ars Combin., in press.[15] M.H.M. Smid, Duadic codes, IEEE Trans. Inform. Theory IT-33 (1987) 432–433.